question on right angle triangle

Question

Let ABC and DBC be two equilateral triangle on the same base BC,a point P is taken on the circle with centre D,radius BD. Show that PA,PB,PC are the sides of a right triangle.

Answer 1

First let's assume $AB=1$ and call $PB=a;PC=b;PA=c$, we have to prove that $a^2+b^2=c^2$. Applying the law of cosines to $PBC$ with respect to angle $\angle BPC=\pi/6$ we get $$a^2+b^2-\sqrt3ab=1$$ Now, call $\angle PCB=y$. Applying the law of cosines to $PBC$ with respect to angle $y$ we get $$ 1+b^2-2b\cos y=a^2$$ hence $\cos y=\frac{1+b^2-a^2}{2b}$. From the law of sines on $PBC$ we also get $\sin y=a/2.$ With these information we can comupte the cosine of angle $\angle PCA=y+\pi/3$: $$cos(y+\pi/3)=\cos y/2-\sin y \sqrt 3/2.$$ Finally we apply the cosine law on $PAC$ with angle $\angle PCA$: \begin{align} c^2&=1+b^2-2b\cos (y+\pi/3) \\ &=1+b^2-2b(\frac{\cos y}{2}-\frac{\sin y \sqrt 3}{2})\\ &=1+b^2-b\big(\frac{1+b^2-a^2}{2b}-\frac{a\sqrt3}{2}\big)\\ &=\frac{1+b^2-a^2}{2}+\frac{ab\sqrt3}{2} \end{align} and plugging in the first equation we find: $$c^2=\frac{1+b^2+a^2}{2}+\frac{a^2+b^2-1}{2}=a^2+b^2.$$